a problem when trying to replace parameters in a lua file - lua

I have never previously worked with lua programming but I have to run a package written in lua language. This package is a command line linux interface and I don't have to know lua programming. In this package I only need to replace some addresses in lua files with my own address and dataset or file names to be able to run the program. But I came across a problem in running a lua file. In this file I replaced the following piece of code:
dofile 'Csv.lua'
local proteinFile = Csv(opt.dataset..".node","r")
local proteinString = proteinFile:readall()
ppFeature = {}
pNumber = {}
for i=1, #proteinString do
local fileName = opt.dataset..'/'..proteinString[i][1]
if file_exists( fileName ) then
local proFile = Csv( fileName, 'r', '\t')
local profile = proFile:readall()
with this one:
dofile 'Csv.lua'
local proteinFile = Csv("/storage/users/ssalmanian/DPPI/myTrain.node","r")
local proteinString = proteinFile:readall()
ppFeature = {}
pNumber = {}
for i=1, #proteinString do
local fileName = '/storage/users/ssalmanian/DPPI/myTrain/proteinString[i][1]'
if file_exists( fileName ) then
local proFile = Csv( fileName, 'r', '\t')
local profile = proFile:readall()
and I got error.
Necessary input files before running the command include a file and a folder with names the same as the -dataset:
A)The suffix of the file is ‘.node’ (myTrain.node). This file has one column which contains names of all proteins.
B) Profile folder with the same name as -dataset (myTrain). This folder contains profiles of all proteins.
The name of the profiles inside this folder is the same as the protein names in ‘.node’ file.
I was wondering if someone could let me know what's wrong with it and which piece of code must be replaced with the following:
local fileName = opt.dataset..'/'..proteinString[i][1]
Thanks.
EDIT:
The package is DPPI package and I tried to run creat_crop.lua code to create two files myTrain_profile_crop_512.t7 and myTrain_number_crop_512.t7 using the following command in a CUDA server :
th creat_crop.lua -dataset myTrain
That creat_crop.lua code is dependent in calling Csv.lua code and needs myTrain.node file and myTrain folder. Here I have added my WinSCP page and the error I get .

Related

Using io.tmpfile() with shell command, ran via io.popen, in Lua?

I'm using Lua in Scite on Windows, but hopefully this is a general Lua question.
Let's say I want to write a temporary string content to a temporary file in Lua - which I want to be eventually read by another program, - and I tried using io.tmpfile():
mytmpfile = assert( io.tmpfile() )
mytmpfile:write( MYTMPTEXT )
mytmpfile:seek("set", 0) -- back to start
print("mytmpfile" .. mytmpfile .. "<<<")
mytmpfile:close()
I like io.tmpfile() because it is noted in https://www.lua.org/pil/21.3.html :
The tmpfile function returns a handle for a temporary file, open in read/write mode. That file is automatically removed (deleted) when your program ends.
However, when I try to print mytmpfile, I get:
C:\Users\ME/sciteLuaFunctions.lua:956: attempt to concatenate a FILE* value (global 'mytmpfile')
>Lua: error occurred while processing command
I got the explanation for that here Re: path for io.tmpfile() ?:
how do I get the path used to generate the temp file created by io.tmpfile()
You can't. The whole point of tmpfile is to give you a file handle without
giving you the file name to avoid race conditions.
And indeed, on some OSes, the file has no name.
So, it will not be possible for me to use the filename of the tmpfile in a command line that should be ran by the OS, as in:
f = io.popen("python myprog.py " .. mytmpfile)
So my questions are:
Would it be somehow possible to specify this tmpfile file handle as the input argument for the externally ran program/script, say in io.popen - instead of using the (non-existing) tmpfile filename?
If above is not possible, what is the next best option (in terms of not having to maintain it, i.e. not having to remember to delete the file) for opening a temporary file in Lua?
You can get a temp filename with os.tmpname.
local n = os.tmpname()
local f = io.open(n, 'w+b')
f:write(....)
f:close()
os.remove(n)
If your purpose is sending some data to a python script, you can also use 'w' mode in popen.
--lua
local f = io.popen(prog, 'w')
f:write(....)
#python
import sys
data = sys.stdin.readline()

Lua Get File Names in folder [duplicate]

I need a list of directory in LUA
Suppose I have a directory path as "C:\Program Files"
I need a list of all the folders in that particular path and how to search any particular folder in that list.
Example
Need a list of all the folder in path "C:\Program Files"
Below are folder name in the above path
test123
test4567
folder 123
folder 456
folder 456 789
Need to get the above in a list and then have to search for a particular string like folder 456 in folder 456 789 only.
Have Tried below code. Something I am missing below:-
local function Loc_Lines( str )
--
local ret= {} -- 0 lines
while str do
local _,_,line,tail= string.find( str, "(.-)\n(.+)" )
table.insert( ret, line or str )
str= tail
Print (str)
end
return ret
end
local function Loc_ShellCommand( cmd )
--
local str= nil
--
local f= io.popen( cmd ) -- no command still returns a handle :(
if f then
str= f:read'*a'
Print(str)
f:close()
end
if str=="" then -- take no output as a failure (we can't tell..)
Print("hi")
str= nil
end
-- Remove terminating linefeed, if any (eases up one-line analysis)
--
if str then
if string.sub( str, -1 ) == '\n' then
str= string.sub( str, 1, -2 )
end
end
return str
end
local function Loc_DirCmd( cmd )
Print(cmd)
local str= Loc_ShellCommand( cmd )
return Loc_Lines(str)
end
local function Loc_DirList( dirname )
local ret= {}
local lookup= {}
local tbl= Loc_DirCmd( "dir /AD /B "..dirname ) -- only dirs
-- Add slash to every dir line
--
for i,v in ipairs(tbl) do
table.insert( ret, v..'\\' )
lookup[v]= true
end
-- Return with forward slashes
--
if true then
for i=1,table.getn(ret) do
ret[i]= string.gsub( ret[i], '\\', '/' )
Print (ret[i])
end
end
return ret
end
Loc_DirList("C:\\Program Files\\")
I hate having to install libraries (especially those that want me to use installer packages to install them). If you're looking for a clean solution for a directory listing on an absolute path in Lua, look no further.
Building on the answer that sylvanaar provided, I created a function that returns an array of all the files for a given directory (absolute path required). This is my preferred implementation, as it works on all my machines.
-- Lua implementation of PHP scandir function
function scandir(directory)
local i, t, popen = 0, {}, io.popen
local pfile = popen('ls -a "'..directory..'"')
for filename in pfile:lines() do
i = i + 1
t[i] = filename
end
pfile:close()
return t
end
If you are using Windows, you'll need to have a bash client installed so that the 'ls' command will work - alternately, you can use the dir command that sylvanaar provided:
'dir "'..directory..'" /b /ad'
Take the easy way, install lfs. Then use the following constructs to find what you need:
require'lfs'
for file in lfs.dir[[C:\Program Files]] do
if lfs.attributes(file,"mode") == "file" then print("found file, "..file)
elseif lfs.attributes(file,"mode")== "directory" then print("found dir, "..file," containing:")
for l in lfs.dir("C:\\Program Files\\"..file) do
print("",l)
end
end
end
notice that a backslash equals [[\]] equals "\\", and that in windows / is also allowed if not used on the cmd itself (correct me if I'm wrong on this one).
for dir in io.popen([[dir "C:\Program Files\" /b /ad]]):lines() do print(dir) end
*For Windows
Outputs:
Adobe
Bitcasa
Bonjour
Business Objects
Common Files
DVD Maker
IIS
Internet Explorer
iPod
iTunes
Java
Microsoft Device Emulator
Microsoft Help Viewer
Microsoft IntelliPoint
Microsoft IntelliType Pro
Microsoft Office
Microsoft SDKs
Microsoft Security Client
Microsoft SQL Server
Microsoft SQL Server Compact Edition
Microsoft Sync Framework
Microsoft Synchronization Services
Microsoft Visual Studio 10.0
Microsoft Visual Studio 9.0
Microsoft.NET
MSBuild
...
Each time through the loop you are given a new folder name. I chose to print it as an example.
I don't like installing libraries either and am working on an embedded device with less memory power then a pc. I found out that using 'ls' command lead to an out of memory. So I created a function that uses 'find' to solve the problem.
This way it was possible to keep memory usage steady and loop all the 30k files.
function dirLookup(dir)
local p = io.popen('find "'..dir..'" -type f') --Open directory look for files, save data in p. By giving '-type f' as parameter, it returns all files.
for file in p:lines() do --Loop through all files
print(file)
end
end
IIRC, getting the directory listing isn't possible with stock Lua. You need to write some glue code yourself, or use LuaFileSystem. The latter is most likely the path of least resistance for you. A quick scan of the docs shows lfs.dir() which will provide you with an iterator you can use to get the directories you are looking for. At that point, you can then do your string comparison to get the specific directories you need.
You also install and use the 'paths' module. Then you can easily do this as follow:
require 'paths'
currentPath = paths.cwd() -- Current working directory
folderNames = {}
for folderName in paths.files(currentPath) do
if folderName:find('$') then
table.insert(folderNames, paths.concat(currentPath, folderName))
end
end
print (folderNames)
-- This will print all folder names
Optionally, you can also look for file names with a specific extension by replacing fileName:find('$') with fileName:find('txt' .. '$')
If you're running on a Unix-based machine you can get a numerically-sorted list of files using the following code:
thePath = '/home/Your_Directory'
local handle = assert(io.popen('ls -1v ' .. thePath))
local allFileNames = string.split(assert(handle:read('*a')), '\n')
print (allFileNames[1]) -- This will print the first file name
The second code also excludes files such as '.' and '..'. So it's good to go!
Don't parse ls, it's evil! Use find with zero-terminated strings instead (on linux):
function scandir(directory)
local i, t = 0, {}
local pfile = assert(io.popen(("find '%s' -maxdepth 1 -print0"):format(directory), 'r'))
local list = pfile:read('*a')
pfile:close()
for filename in s:gmatch('[^\0]+')
i = i + 1
t[i] = filename
end
return t
end
WARNING: however, as an acceped answer this apporach could be exploited if directory name contain ' in it. Only one safe solution is to use lfs or other special library.
Few fixes of val says Reinstate Monica solution:
function scandir(directory)
local pfile = assert(io.popen(("find '%s' -mindepth 1 -maxdepth 1 -type d -printf '%%f\\0'"):format(directory), 'r'))
local list = pfile:read('*a')
pfile:close()
local folders = {}
for filename in string.gmatch(list, '[^%z]+') do
table.insert(folders, filename)
end
return folders
end
Now it filters by folders, excludes dir itself and prints only names.

How to load host.conf file variables in lua script

I need to load configuration variables from .conf file in lua script, and use those variables to connect to a database. I have tried using:
require "host.conf"
loadfile("host.conf") - error with unexpected token '#'
os.execute("pathToConfFile/host.lua") - and I have created a lua host file with variables in bash shell
io.popen("host.conf") etc..
None of these solutions are valid.
Is there a way to use the existing host.conf file in lua, and avoid the unexpected token error?
Thank you for your suggestions.
local original = io .open('host.conf')
local hostconf = {} -- copy contents into Lua table
for line in original :lines() do
table .insert( hostconf, line )
end ; io .close( original )
print( hostconf[1] ) -- prints line 1
You haven't specified what format your host.conf comes in, but you'll likely want to parse it better than just throwing contents in a list. Perhaps splitting each line into head / tail, based upon a delimiter ( comma, space, whatever you have between variable & value )
Thank you to everyone who helped out. My question wasn't precise and I had more to learn before I have asked and sorry about that. This is what I used to solve the problem.
local open = io.open
local function read_file(path)
local file = open(path, "r")
if not file then return nil end
local content = file:read "*a" -- *a or *all reads the whole file
local lines = {}
for line in io.lines(path) do
--print(line);
if(line:find(var)~=nil)then
local varStart=string.len(var)+2
local varEnd=string.len(line)
var=string.sub(line,varStart,varEnd)
print(var);
end
--repeat for every line
end
file:close()
return lines;
end
local fileContent = read_file("path");

How to copy latest file from a directory using lua

I am trying to copy only latest file from a directory using lua file.
Latest file means : depends on modified time/created time.
How can i do this?
Referring to this question: How can I get last modified timestamp in Lua
You might be able to leverage the io.popen function to execute a shell command to get the name of the file. It seems like there's no builtin function that exposes filesystem metadata or stats. Something like this might work:
local name_handle = io.popen("ls -t1 | head -n 1")
local filename = name_handle:read()
I'm not familiar with Lua, but perhaps this helps. I imagine that once you have the name of the newest file you can use the other IO functions to do the copying.
local function get_last_file_name(directory)
local command = 'dir /a-d /o-d /tw /b "'..directory..'" 2>nul:'
-- /tw for last modified file
-- /tc for last created file
local pipe = io.popen(command)
local file_name = pipe:read()
pipe:close()
return file_name
end
local directory = [[C:\path\to\your\directory\]]
local file_name = get_last_file_name(directory)
if file_name then
print(file_name)
-- read the last file
local file = io.open(directory..file_name)
local content = file:read"*a"
file:close()
-- print content of the last file
print(content)
else
print"Directory is empty"
end

How to change local file path into ARGF for automatically input in Rails

I have my local database file in public folder. I would like to open it in my Rails app without running it manually in a terminal. The command requires ARGF (Gem bio) which will work by
In terminal,
>> Ruby thiscode.rb database.fasta
And in .rb file (Model),
input_seq = ARGF.read
my_aaseq = Bio::Sequence::AA.new(input_seq)
Or
fff = Bio::FlatFile.new(Bio::FastaFormat, ARGF)
Then the .fasta file could be iterated.
I did Rails.root.join("public", "database.fasta") and .to_s for input as ARGF but it doesn't work. How can I create ARGF here from local file?
Additional Detail
I checked fff variable to see how different between the working one (by ARGF) and not-working one (by local file). In short,
the working one: #io=ARGF and #path=database.fasta
the not-working one: #io = full file path from root join, and #path=nil.
Here are the full differences
work
<Bio::FlatFile:0x007fdfb104ef78 #stream=#<Bio::FlatFile::BufferedInputStream:0x007fdfb207a7a8 #io=ARGF, #path="database.fasta", #buffer="">, #dbclass=Bio::FastaFormat, #splitter=#<Bio::FlatFile::Splitter::Default:0x007fdfb1900158 #dbclass=Bio::FastaFormat, #stream=#<Bio::FlatFile::BufferedInputStream:0x007fdfb207a7a8 #io=ARGF, #path="database.fasta", #buffer="">, #entry_pos_flag=nil, #delimiter="\n>", #header=nil, #delimiter_overrun=1>, #skip_leader_mode=:firsttime, #firsttime_flag=true, #raw=false>
not work
<Bio::FlatFile:0x007fc04ae033c0 #stream=#<Bio::FlatFile::BufferedInputStream:0x007fc0511d3e18 #io="/Users/folder/NAcamp/pt_app/rails_app/Peptool/public/database.fasta", #path=nil, #buffer="">, #dbclass=Bio::FastaFormat, #splitter=#<Bio::FlatFile::Splitter::Default:0x007fc04bdaaf08 #dbclass=Bio::FastaFormat, #stream=#<Bio::FlatFile::BufferedInputStream:0x007fc0511d3e18 #io="/Users/folder/NAcamp/pt_app/rails_app/Peptool/public/database.fasta", #path=nil, #buffer="">, #entry_pos_flag=nil, #delimiter="\n>", #header=nil, #delimiter_overrun=1>, #skip_leader_mode=:firsttime, #firsttime_flag=true, #raw=false>

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